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For a prime number $p$ and a positive integer $n$, the prime valuation is the exponent of $p$ in the prime factorization of $n$, equivalentlyThus $v_2(360)=3$ because $8$ divides $360$ but $16$ does not; $v_3(360)=2$, $v_5(360)=1$ and $v_7(360)=0$. In particular $v_p(1)=0$. Multiplication adds prime valuations, $v_p(ab)=v_p(a)+v_p(b)$, and the fundamental theorem of arithmetic makes comparison of prime valuations an exact test for divisibility between positive integers. This definition extends to a nonzero integer by taking its absolute value; for zero one commonly sets $v_p(0)=+\infty$.

A sum whose resulting value is a prime number. This is a condition on the value, rather than a requirement that the summation index run over prime numbers. Counting subsets with a prime-valued sum can use subset sum counts followed by a primality filter on the possible totals.

For a prime number $p$, a p-valuation is a separated filtration on a group such that every $x\ne 1$ satisfiesA group equipped with one is called a p-valued group.

Quadratic reciprocity compares two questions: is the prime number $p$ a quadratic residue modulo $q$, and is $q$ a quadratic residue modulo $p$? Let $p,q$ be distinct odd primes. In terms of the Legendre symbol,The two Legendre symbols have opposite signs exactly when both primes are $3$ modulo $4$; otherwise they are equal. Their values are nonzero because the primes are distinct. Equal signs can mean two quadratic residues or two quadratic nonresidues: reciprocity compares the answers rather than declaring either answer positive.^[ref]

For a direct computation, the nonzero quadratic residues modulo $3,5,7$ are respectively $\{1\}$, $\{1,4\}$, and $\{1,2,4\}$. Squaring just $1,\dots ,(p-1)/2$ suffices because $x$ and $p-x$ have the same square modulo an odd prime $p$.The pair $3,7$ has a sign flip, whereas $5,7$ has two negative signs. These small quadratic residue tables make the distinction tangible.

The first supplementary law for quadratic reciprocity handles numerator $-1$, and the second supplementary law for quadratic reciprocity handles numerator $2$:For an odd prime $p$, $-1$ is a quadratic residue exactly when $p\equiv 1(\mathrm{mod}4)$, while $2$ is a quadratic residue exactly when $p\equiv 1,7(\mathrm{mod}8)$. The remaining odd congruence classes modulo $8$, namely $3,5$, make $2$ a quadratic nonresidue. These laws fill the cases excluded from swapping distinct odd primes.^[ref]

For an odd prime $p$, a nonzero congruence class $a$ is a quadratic residue when $a\equiv x^2(\mathrm{mod}p)$ for some integer $x$. Here "quadratic residue" excludes zero: zero is a square of a number, but is neither a quadratic residue nor a quadratic nonresidue under this convention. Some treatments include zero among quadratic residues; the distinction matters when counting them or discussing their subgroup.^[ref]

Modulo seven, squaring the six nonzero congruence classes givesThus the set of quadratic residues is $Q_7=\{1,2,4\}$, and the quadratic nonresidues are $\{3,5,6\}$. In particular, $x^2\equiv 2(\mathrm{mod}7)$ has exactly the two solutions $x\equiv 3,4(\mathrm{mod}7)$: $3^2=9\equiv 2$ and $4^2=16\equiv 2$. Modulo eleven the quadratic residues are instead $Q_{11}=\{1,3,4,5,9\}$. Two is therefore a quadratic residue modulo seven and a quadratic nonresidue modulo eleven: being a quadratic residue depends on the modulus.

Why do the square roots come in pairs? If $x^2\equiv y^2(\mathrm{mod}p)$, the prime number $p$ divides $(x-y)(x+y)$, so it divides one factor: $y\equiv x$ or $y\equiv -x(\mathrm{mod}p)$. For nonzero $x$ these are distinct, because $p$ is an odd prime. Every nonempty nonzero fiber of a function of $s(x)=x^2$ therefore contains exactly two inputs. Pairing the $p-1$ nonzero congruence classes gives exactly $(p-1)/2$ quadratic residues and $(p-1)/2$ quadratic nonresidues. Including zero gives $(p+1)/2$ different squares, with one rather than two square roots for zero.

The quadratic residues form a subgroup of the multiplicative group of a finite field ${\mathbb{F}}_p^{\times }$, with index of a subgroup two. Indeed, one is a square, $(u^2)(v^2)=(uv{)}^2$, and $(u^2{)}^{-1}=(u^{-1}{)}^2$; the inverses exist because the inputs are nonzero in the prime field. Equivalently, the squaring group homomorphism has kernel of a group homomorphism $\{1,-1\}$ and image of a function $Q_p$. The apparent loss of a sign in squaring becomes a precise division of the nonzero arithmetic into two equal parts.

This structure turns a search for square roots into a yes-or-no test: Euler's criterion detects quadratic residues through modular exponentiation, and the Legendre symbol records the result as a sign. For a general integer modulus, the modular congruence $x^2\equiv a(\mathrm{mod}n)$ still makes sense, but the two-root count and the prime field argument need not survive. For example, $x^2\equiv 1(\mathrm{mod}15)$ has four solutions $1,4,11,14$; the Chinese remainder theorem separates the sign choices modulo three and five.

The radical of an integer $n>0$ is the product of its distinct prime factors, each included exactly once. If the prime factorization is $n=p_1^{a_1}\cdots p_r^{a_r}$ with distinct prime numbers $p_i$ and positive integer exponents $a_i$, thenThus the radical of an integer remembers which prime factors occur and forgets their prime valuations. At $n=1$, the empty product gives $\mathrm{rad}(1)=1$. The domain of a function here is the positive integers; zero is excluded. ^[ref]

For example, $72=2^3\cdot 3^2$ has distinct prime factors $2,3$, so $\mathrm{rad}(72)=2\cdot 3=6$. Similarly,using the prime factorizations $30=2\cdot 3\cdot 5$, $32=2^5$ and $36=2^2\cdot 3^2$. The prime omega function counts the retained prime factors, while the total number of prime factors counts their occurrences before repetitions are removed.

The radical of an integer is its largest squarefree divisor. Indeed, a squarefree divisor chooses at most one copy of each prime factor, so it divides $\mathrm{rad}(n)$; the radical of an integer itself is a squarefree divisor. Consequently,The last statement follows because every nonzero prime valuation is then one. It includes $n=1$, a squarefree integer. Applying the radical of an integer twice also changes nothing: $\mathrm{rad}(\mathrm{rad}(n))=\mathrm{rad}(n)$.

For any positive integers $a,b$, the exact product rule isTo prove it, a prime factor dividing just one input occurs once in the numerator and not in the denominator. A prime factor dividing both occurs twice in the numerator and once in the denominator, since it divides their greatest common divisor. Each prime factor of $ab$ therefore remains exactly once. For coprime integers, the denominator is $\mathrm{rad}(1)=1$, proving the radical of an integer is a multiplicative function. It is not a completely multiplicative function: $\mathrm{rad}(4)=2\ne 4=\mathrm{rad}(2{)}^2$.

For a worked example with shared prime factors, take $a=12$ and $b=18$. Their greatest common divisor is $6$, and $ab=216=2^3\cdot 3^3$, soThis explains why prime support of an integer is useful: questions about the presence of prime factors can discard their multiplicities. For example, the squarefree divisors of $72$ are exactly the integer divisors of $\mathrm{rad}(72)=6$, namely $1,2,3,6$. Also $\gcd (m,n)=1$ if and only if $\gcd (m,\mathrm{rad}(n))=1$, because coprimality depends only on shared prime factors.

A further characterization uses mutual divisibility of powers: for positive integers $a,b$,If their prime supports of an integer agree, sufficiently large $k,\ell$ make all the required prime valuations large enough. Conversely, these two divisibility relations force each prime factor of either input to divide the other. For $72$ and $36$, one can take $k=2$ and $\ell =1$: $72\mid 36^2$ and $36\mid 72$.

For a prime number $p$, reducing every entry modulo $p$ defines a group homomorphism from an integral matrix group to the corresponding matrix group over ${\mathbb{F}}_p$. The target is finite. If an integral matrix $M$ is not the identity, choosing $p$ not to divide one nonzero entry of $M-I$ ensures that its reduction is also nonidentity.

A repdigit is a positive integer whose base-$b$ expansion repeats one nonzero digit $d$. For $b\ge 2$, $1\le d<b$ and $n\ge 1$, the finite geometric series gives $R_n=d(1+b+\cdots +b^{n-1})=d(b^n-1)/(b-1)$. For example, $5(10^n-1)/9$ has $n$ decimal digits, all equal to five.

Every prime number not dividing $b$ divides infinitely many such repdigits. If it divides $d$, every length works. Otherwise, if it does not divide $b-1$, the Fermat little theorem gives divisibility for lengths that are multiples of $p-1$, and division by $b-1$ is justified by its modular inverse. If $p\mid b-1$, the sum is congruent to $dn$ modulo $p$, so lengths divisible by $p$ work instead. A prime dividing $b$ but not $d$ divides none, since $R_n\equiv d(\mathrm{mod}p)$. This distinction prevents invalid cancellation of the geometric-sum denominator.

Let $\alpha$ be an algebraic irrational number, let $S$ be a finite set of primes, and let $\epsilon >0$. There are only finitely many reduced fractions $p/q$ whose denominator is S-smooth number and which satisfyThis is the denominator-restricted consequence of the finite-place Schmidt subspace theorem.

For a prime number $p$ and $\alpha =\sqrt[p]{p}$, the polynomial $X^p-p$ is Eisenstein at $p$, andThe discriminant-index formula for an integral lattice reduces the proof to the prime $p$. At the unique prime ideal above $p$, one has $v(\alpha )=1$ and $v(p)=p$, which rules out an integral element $g(\alpha )/p$ with $g\notin p\mathbb{Z}[X]$.

The Second Chebyshev function is the partial sum of the Von Mangoldt function up to a real cutoff $x\ge 0$:For example, $\psi (3)=\log 6$, $\psi (4)=\log 12$, and $\psi (4.5)=\log 12$. At $4=2^2$ the jump has size $\log 2$; between $4$ and $5$ no new prime power greater than one enters the sum. In contrast, the Chebyshev theta function stays at $\vartheta (4)=\log 6$.

There is also an exact connection with the least common multiple. For $N=\lfloor x\rfloor \ge 1$, each prime number $p\le N$ occurs to the highest exponent $k$ for which $p^k\le N$, both in that least common multiple and in the accumulated Von Mangoldt function weights. ThusThis explains the $\log 2520$ value at $x=10$: the least common multiple is $2^3\cdot 3^2\cdot 5\cdot 7$. The Chebyshev theta function instead gives the natural logarithm of the primorial $2\cdot 3\cdot 5\cdot 7=210$. The additional prime powers retain the highest required multiplicities.

Why is this useful for the distribution of prime numbers? The Prime number theorem has the equivalent formswhere $\pi$ is the prime-counting function and $\sim$ denotes asymptotic equivalence. Thus an arithmetic function defined using prime factorization has average weight tending to one over long initial intervals of the positive integers. These are substantive asymptotic equivalences, not consequences of the definitions or the Von Mangoldt divisor identity alone. ^[ref]

The elementary reason the two Chebyshev functions have the same leading behavior is that their difference is small compared with $x$. Every additional prime power has base $p\le \sqrt{x}$, and for each such prime number its total extra weight is at most $\log x$. There are at most $\lfloor \sqrt{x}\rfloor$ such bases, soThis transfers $\psi (x)\sim x$ to $\vartheta (x)\sim x$, or conversely, once either is established. Partial summation then relates the Chebyshev theta function to the prime-counting function. The estimate controls the extra prime powers; proving the leading value itself is the harder Prime number theorem problem.

A semiprime is a composite number that is a product of exactly two prime numbers, counted with multiplicity. For example, $6=2\cdot 3$ and $9=3^2$ are semiprimes. Distinct factors give $\phi (pq)=(p-1)(q-1)$ for the Euler totient function, whereas equal factors give $\phi (p^2)=p(p-1)$. This distinction matters when constructing a uniform coprime state for a semiprime.

The sieve of Eratosthenes finds all prime numbers at most an integer $N\ge 2$. Start with the integers $2,\dots ,N$ marked as candidates. For each surviving candidate $p$ in increasing order, mark the multiples $p^2,p^2+p,\dots$ up to $N$ as composite numbers. Stop processing candidates once $p^2>N$: the surviving candidates are exactly the prime numbers. Starting at $p^2$ saves work because any smaller proper multiple of $p$ already has a smaller prime factor. ^[ref].

For a complete run with $N=60$, the only prime numbers at most $\sqrt{60}$ are $2,3,5,7$, since $7^2\le 60<8^2$. Exclude one before starting; it is not a prime number. The four marking passes are:

There are $59-29-9-3-1=17$ surviving prime numbers:The next surviving candidate is $11$, and $11^2>60$, so no further marking pass is necessary. A prime number used for marking remains a candidate: starting at $p^2$ never marks $p$ itself.

The saving from starting at $p^2$ is justified by the smaller multiplier: if $pk<p^2$ with $2\le k<p$, then $k$ has a prime factor $q<p$, so $q$ has already marked $pk$. For $p=7$, the earlier multiples $14,21,28,35,42$ have already been marked using $2,3$ or $5$. The first new possibility is $49$.

For $N=100$, only the prime numbers $2,3,5,7$ need marking passes. They remove all the composite numbers, leaving 25 prime numbers. The diagram colors each composite number by its smallest prime factor, so it shows which marking pass removes it first. The excluded integer one is shown separately.

Why does the algorithm work? A surviving candidate cannot be a composite number, since its smaller prime factor would already have removed it. Conversely, marking proper multiples never removes a prime number. Every composite number at most $N$ has a prime factor at most $\sqrt{N}$, so no later pass is needed. The sieve of Eratosthenes is useful when an entire bounded list is wanted; primality testing of one large integer is a different computational task.

The historical name is supported by a surviving primary account: Nicomachus, in Book I, chapter 13 of Introduction to Arithmetic, names Eratosthenes and describes separating prime numbers from composite numbers by marking multiples among the odd integers beginning with three. That passage documents the sieve tradition; the bounded algorithm above and its $p^2$ starting rule are the formulation explained here, rather than instructions quoted from Euclid's Elements. ^[ref]

For the displayed integral equation, reduce its polynomial modulo a prime number $p$. Singular points satisfy $y^2-x^3-c=-3x^2=2y=0$. For $p>3$, they exist exactly when $p\mid c$, and the sole point is $(0,0)$. At $p=2$, the sole point over ${\mathbb{F}}_2$ is $(0,\overline{c})$; at $p=3$, it is $(-\overline{c},0)$ because cubing is the identity on ${\mathbb{F}}_3$. For $c=17$, this gives primes $2,3,17$ and points $(0,1),(1,0),(0,0)$ respectively. This concerns the specified plane equation; intrinsic good or bad reduction of an elliptic curve can require choosing a different integral model, especially at primes two and three.

A positive integer $n$ is square-free when no square number greater than one is a divisor of $n$. It suffices to exclude squares of prime numbers; equivalently, every prime valuation is zero or one:Thus the prime factorization of a square-free integer is a product of distinct prime factors, with no repetitions. One is included: its prime factorization is an empty product and all its prime valuations are zero. These equivalent tests are stated in Grinberg's notes. ^[ref]

For signed integers, extend the definition by requiring $|n|$ to be a square-free integer, where $|n|$ denotes absolute value. Hence $-30$ and both $1,-1$ qualify. Zero does not: every prime number has its square as a divisor of zero. The Möbius function and radical of an integer below use their existing positive integer domains, so apply them to $|n|$ when handling a negative integer.

For example, $30=2\cdot 3\cdot 5$ is a square-free integer, whereas $12=2^2\cdot 3$ is not, because $4\mid 12$ and $v_2(12)=2$. Being a composite number causes no problem: $6,10,15$ are all square-free integers. A square-free integer need not be coprime to another integer: $6$ and $10$ are both square-free integers but have greatest common divisor two. In fact the only signed integers coprime to every integer are $1,-1$, since $\gcd (n,n)=|n|$.

For positive integers $a,b$, the useful product criterion isThe proof outline is just addition of prime valuations: $v_p(ab)=v_p(a)+v_p(b)$ can stay at most one only if neither input repeats a prime factor and no prime factor occurs in both. Thus $6\cdot 5=30$ is a square-free integer, but $6\cdot 10=60$ is not. An equivalent characterization is that every factorization $n=ab$ into positive integers has coprime factors. If $p^2\mid n$, the factorization $n=p(n/p)$ supplies two factors sharing $p$; conversely, a shared prime factor forces its square to divide $n$. For $30$, the unordered factor pairs $(1,30),(2,15),(3,10),(5,6)$ all have greatest common divisor one; for $12$, the pair $(2,6)$ fails.

Two existing arithmetic functions turn the definition into practical tests:The radical of an integer keeps one copy of each prime factor, so it changes precisely the integers with a repeated prime factor. Its definition is given in Evertse's exercise sheet. ^[ref] The Möbius function assigns $(-1{)}^r$ to a square-free integer with $r$ distinct prime factors and zero otherwise; squaring forgets the sign. Consequently ${\sum }_{n=1}^{60}\mu (n{)}^2=37$ counts exactly the green cells. This connects a restriction on repeated prime factors with weighted counting and inclusion-exclusion. ^[ref]

If $S$ is a finite set of prime numbers, every S-smooth number has a unique expression $n=a{b}^2$, where $a$ is squarefree and its prime factors lie in $S$. There are at most $2^{|S|}$ choices for $a$ and at most $\lfloor \sqrt{x}\rfloor$ choices for $b$ when $n\le x$. Consequently $f_S(x)\le 2^{|S|}\sqrt{x}$.

A number $p^2{q}^3$ for distinct prime numbers $p,q$. The ordering of the two primes matters because their exponents differ. Unique prime factorization identifies the ordered pair from the resulting integer.

For a finite set $S$ of primes, an integer is $S$-smooth when every prime dividing it belongs to $S$. Bounding the largest prime factor is equivalent to restricting the integer to be $S$-smooth for a suitable finite $S$.

Every integer $n\ge 2$ has a prime factorization. The induction base case $2$ is prime. For a larger integer, either it is prime or it is $ab$, with $2\le a,b<n$. The induction hypothesis supplies prime factorizations of both smaller integer divisors; multiplying them gives one for $n$. This proves existence, while uniqueness requires the additional argument under the fundamental theorem of arithmetic.^[ref]

For the chosen number $84$, one possible recursive decomposition isThe step at $84$ needs results for $6$ and $14$, rather than a result for $83$. Each composite number in this factorization tree splits into strictly smaller positive integers, so recursive descent terminates at prime numbers. The tree describes one construction; another choice of integer divisors can give a differently shaped factorization tree.

The same pattern explains the usefulness of strong induction for algorithm correctness. When a recursive procedure returns answers for smaller inputs, the induction hypothesis can justify those answers; the remaining task is to justify how the current call combines them. Strictly decreasing a nonnegative integer size measure establishes termination. Correctness also requires each call's conditions and its combination rule; termination alone does not make an algorithm correct.

Let $N=pq$ for distinct odd prime numbers $p,q$ with $\gcd (p-1,q-1)=10$. Exactly $100$ unit bases make $N$ a Fermat pseudoprime, and exactly $50$ of these make it a strong pseudoprime. The Fermat bases form $C_{10}\times C_{10}$ under the Chinese remainder theorem for unit groups. If $N-1=2^{s}m$, then $5\mid m$ and $m$ is odd, so the $m$th-power map sends five elements of each $C_{10}$ to $1$ and five to $-1$. The strong condition requires equal signs in both components, giving $5^2+5^2=50$.